Genady

@dqqd, this analysis in Wikipedia is quite clear: Variable-mass system - Wikipedia

I disagree: the force has not been redefined. In the formula \(F=\frac {dP}{dt}\), \(P\) is the momentum of the entire system. If the system is composed of parts, \(P=P_1+P_2+...\), then \(F=\frac {d(P_1+P_2+...)}{dt}=m_1 \frac {dv_1}{dt}+v_1 \frac {dm_1}{dt}+m_2 \frac {dv_2}{dt}+v_2 \frac {dm_2}{dt}+...\), where no mass crosses the boundary of the composed system.

Force, however, does not depend on the inertial frame you choose.

Morning with the family

But in the OP, v is not "the velocity of the ejected mass relative to the bulk mass," but rather (my emphasis), It depends on reference frame.

There is no rocket in my scenario. My scenario demonstrates that by applying the quoted formula wrongly, we get a ridiculous result that a force on the bucket is proportional to velocity with which an observer runs by it.

Remove the bucket and consider a ball of water in free fall. The molecules leave in all directions. The total force on the ball is zero.

Consider a bucket of water with mass \(m\) and the water evaporating at rate \(\frac {dm}{dt}\). No force is applied to the bucket. Consider now the same bucket in a reference frame where it moves with velocity \(v\). If we apply the formula "F=dp/dt=mdv/dt+vdm/dt" in this reference frame, we get a ridiculous result that there is a force \(F=v \frac {dm}{dt}\) on the bucket. Conclusion: this is a wrong way to apply the above formula.

Local inhomogeneities and anisotropies do not matter on cosmological scales. The cosmological principle's assumption is that on some large scale, currently about 100+ Mpc AFAIK, the universe is homogeneous and isotropic. All manifestations of the DE so far are on such scales.

If it does, the cosmological principle needs to be reconsidered.

That is perhaps what he did and now tries to convince himself that he should stop feeling guilty. In his own words,

The Lizard.

Breakfast

They cannot land as anything else just by the fact that they are primes.

That will be good. I don't want to die young, healthy, and/or dirty.

Yes! +1.

I disagree. Computational methods are math by itself which are applied to math of the physical theories. E.g., FEM.

In fact, simulations are MORE math. Given equations that cannot be solved exactly, developing a working method for solving them numerically is a nontrivial math itself.

If I'm not mistaken, there is an error in the last part:

Scientists discover the world’s largest coral — so big it can be seen from space | CNN This is cool, but the following video in the story made me, a retired SCUBA instructor, cringe. SCUBA divers / marine biologists, do you see what is wrong with this picture? https://media.cnn.com/api/v1/loops/stellar/prod/video-world-s-largest-coral-4.mp4?c=original

Multiplication example. Take 13 jars. Put 19 beans in each jar. Take out all the beans and count them. The number, 247, is a result of multiplying 13 by 19. Inverse multiplication example. Take 247 beans. Put them one-by-one in 13 jars, one bean in each jar at a time, until all the beans are in the jars. Take out beans from one of the jars and count them. The number, 19, is a result of the inverse operation (aka dividing 247 by 13.)

I think that the fruit tree metaphor is wrong. It assumes that physicists are standing now on the same ground level as Einstein, Heisenberg, Feynman et al were standing on, while the branches are farther away. It is not so. The branches reached by Einstein, Heisenberg, Feynman et al are the new ground level, and the "higher" branches are not necessarily farther away.

AFAIK, Galileo worked freely and productively for about two decades (and even under Pope's patronage) until the Inquisition decided to stop it.

Yes. It is easy to show that such a polynomial does not exist for not integer n. So, yes, the question can be clarified: n and all coefficients of the polynomial being integers. It is not needed for this puzzle. Here is almost complete proof:

I mean that. +1. As a bonus exercise, prove that it is the only answer.